NCERT Solutions For Class 11 Maths chapter-6 Linear Inequalities Miscellaneous Exercise

NCERT Solutions For Class 11 Maths chapter-6 Linear Inequalities Miscellaneous Exercise

NCERT Solutions for Class 11 Maths Chapter-6 Linear Inequalities

Academic team of Physics Wallah developed step by step NCERT Solutions For Class 11 Maths Chapter-6 Linear Inequalities Miscellaneous Exercise according to recommen dations and Guideline of CBSE. You can download solution of all chapters from Physics Wallah NCERT solutions of class 11.

NCERT Solutions for Class 11 Maths Miscellaneous Exercise

Question 1. Solve the inequality 2 ≤ 3x – 4 ≤ 5

Solution :
2 ≤ 3x – 4 ≤ 5

⇒ 2 + 4 ≤ 3x – 4 + 4 ≤ 5 + 4

⇒ 6 ≤ 3x ≤ 9

⇒ 2 ≤ x ≤ 3

Thus, all the real numbers, x, which are greater than or equal to 2 but less than or equal to 3, are the solutions of the given inequality. The solution set for the given inequalityis [2, 3].

Question 2. Solve the inequality 6 ≤ –3(2x – 4) < 12

Solution :
6 ≤ – 3(2x – 4) < 12

⇒ 2 ≤ –(2x – 4) < 4

⇒ –2 ≥ 2x – 4 > –4

⇒ 4 – 2 ≥ 2x > 4 – 4

⇒ 2 ≥ 2x > 0

⇒1 ≥ x > 0

Thus, the solution set for the given inequalityis (0, 1].

Question 3.

Solution :
Given:

As a result, the set of solutions for the given inequality is [−4,2].

Question 4.

Solution :
Given:

⇒  −12x−12<30−12

⇒ −75<3(x−2)⩽0

⇒ −25<x−2⩽0

⇒ −25+2<x⩽2

⇒ −23<x⩽2

As a result, the set of solutions for the given inequality is (-23, 2]

Question 5.

Solution :

Question 6.

Solution :

Question 7. Solve the inequalities and represent the solution graphically on number line: 5x + 1 > –24, 5x – 1 < 24

Solution :
5x + 1 > –24

⇒ 5x > –25

⇒ x > –5 … (1)

5x – 1 < 24

⇒ 5x < 25

⇒ x < 5 … (2)

From (1) and (2), it can be concluded that the solution set for the given system of inequalities is (–5, 5). The solution of the given system of inequalities can be represented on number line as

Question 8. Solve the inequalities and represent the solution graphically on number line: 2(x – 1) < x + 5, 3(x + 2) > 2 – x

Solution :
2(x – 1) < x + 5

⇒ 2x – 2 < x + 5

⇒ 2x – x < 5 + 2

⇒ x < 7 … (1)

3(x + 2) > 2 – x

⇒ 3x + 6 > 2 – x

⇒ 3x + x > 2 – 6

⇒ 4x > – 4

⇒ x > – 1 … (2)

From (1) and (2), it can be concluded that the solution set for the given system of inequalities is (–1, 7). The solution of the given system of inequalities can be represented on number line as

Question 9. Solve the following inequalities and represent the solution graphically on number line:

3x – 7 > 2(x – 6), 6 – x > 11 – 2x

Solution :
3x – 7 > 2(x – 6)

⇒ 3x – 7 > 2x – 12

⇒ 3x – 2x > – 12 + 7

⇒ x > –5 … (1)

6 – x > 11 – 2x

⇒ –x + 2x > 11 – 6

⇒ x > 5 … (2)

From (1) and (2), it can be concluded that the solution set for the given system of inequalities is(5, ). The solution of the given system of inequalities can be represented on number line

Question 10. Solve the inequalities and represent the solution graphically on number line: 5 (2x – 7) – 3 (2x + 3) ≤ 0 , 2x + 19 ≤ 6x + 47 .

Solution :
5(2x – 7) – 3(2x + 3) ≤ 0

⇒ 10x – 35 – 6x – 9 ≤ 0

⇒ 4x – 44 ≤ 0

⇒ 4x ≤ 44

⇒ x ≤ 11 … (1)

2x + 19 ≤ 6x + 47

⇒ 19 – 47 ≤ 6x – 2x

⇒ –28 ≤ 4x

⇒ –7 ≤ x … (2)

From (1) and (2), it can be concluded that the solution set for the given system of inequalities is [–7, 11].

The solution of the given system of inequalities can be represented on number line as

Question 11. A solution is to be kept between 68°F and 77°F. What is the range of temperature in degree Celsius (C) if the Celsius / Fahrenheit (F) convension formula is given by ?

Solution :
Given 68°F and 77°F.

Question 12. A solution of 8% boric acid is to be diluted by adding a 2% boric acid solution to it. The resulting mixture is to be more than 4% but less than 6% boric acid. If we have 640 litres of the 8% solution, how many litres of the 2% solution will have to be added?

Solution :
Let x litres of 2% boric acid solution is required to be added.

Then, total mixture = (x + 640) litres

This resulting mixture is to be more than 4% but less than 6% boric acid.

∴2%x + 8% of 640 > 4% of (x + 640)

And, 2% x + 8% of 640 < 6% of (x + 640)

2%x + 8% of 640 > 4% of (x + 640)

⇒ 2x + 5120 > 4x + 2560

⇒ 5120 – 2560 > 4x – 2x

⇒ 5120 – 2560 > 2x

⇒ 2560 > 2x

⇒ 1280 > x

2% x + 8% of 640 < 6% of (x + 640)

⇒ 2x + 5120 < 6x + 3840

⇒ 5120 – 3840 < 6x – 2x

⇒ 1280 < 4x

⇒ 320 < x

∴320 < x < 1280

Thus, the number of litres of 2% of boric acid solution that is to be added will have to be more than 320 litres but less than 1280 litres.

Question 13. How many liters of water will have to be added to 1125 liters of the 45% solution of acid so that the resulting mixture will contain more than 25% but less than 30% acid content?

Solution :
Allow for the addition of x litres of water. The entire mixture is then calculated = (x + 1125) litres It is clear that the amount of acid in the final mixture is excessive. 45% of 1125 litres. The resulting mixture will have a higher concentration of 25% but less than 30% acid content.

Therefore, minimum 562.5 liters and maximum 900 liters of water need to be added.

Question 14. IQ of a person is given by the formula IQ = Where MA is mental age and CA is chronological age. If 80 ≤ IQ ≤ 140 for a group of 12 years old children, find the range of their mental age.

Solution :